"In the expansion of (2-ax)^5 the coefficient of x^2 is 5. Find the value of a".

I know I have to use the binomial theorem to help me, but I don't know how.

I got it. :)

How did u get it?

According to the binomial theorem, the term with x^2 would be the third term. I plugged my known values into the "formula" for the third term, and made it equal to 5. Then I solved for a. Sorry, my explanation is kind of vague.

Cool, I appreciate you responding to my question.

To find the value of a in the expansion of (2-ax)^5, you can indeed use the Binomial Theorem. The Binomial Theorem states that for any positive integer n:

(x + y)^n = C(n, 0)x^n y^0 + C(n, 1)x^(n-1)y^1 + C(n, 2)x^(n-2)y^2 + ... + C(n, n-1)x^1 y^(n-1) + C(n, n)x^0 y^n

Where C(n, k) represents the binomial coefficient, which can be calculated as:

C(n, k) = n! / (k!(n-k)!)

In this case, we want to find the coefficient of x^2, which means we want to find the term with k = 2 in the expansion of (2-ax)^5. According to the Binomial Theorem, this term can be calculated as:

C(5, 2)(2)^(5-2)(-ax)^2 = 5(2)^3(-a)^2 x^2

Simplifying further, we have:

5 * 8 * a^2 * x^2 = 40a^2 x^2

Now we know that the coefficient of x^2 is 5, so we can set up the equation:

40a^2 = 5

To find the value of a, we can solve for a by dividing both sides of the equation by 40 and then taking the square root of both sides:

a^2 = 5/40
a^2 = 1/8
a = ± √(1/8)

So, the value of a can be either a = √(1/8) or a = -√(1/8).